regular expression or string (nevermind) format
Resolved,
I can use the string to a fractional number!
Hello
I have an array of strings in a multicolumn list box 2D, and I want to check if each value is a valid number, numbers can be one and can have negative values. Anyone know what regular expression, I need to check it out, or there at - it a still better way?
Best regards
Thijs
solved!
Tags: NI Software
Similar Questions
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Multiline - Regular Expression Match string
I'm trying to understand the format of a regular expression to pull select off multi-line string lines and fill in these lines as the individual elements of an array of strings using regular expressions to Match. The total length of the multiline string may vary as well as the text in the string. The string can contain letters, numbers and special characters. I've attached an example VI. In the example of VI, I want to only return lines starting by "device #" in the table. The number of lines starting by "device #" can vary, but I want to capture them all.
Or is there a better functioning to be used instead of the corresponding regular Expression that will give me the desired result?
aaronb wrote:
I'm trying to understand the format of a regular expression to pull select off multi-line string lines and fill in these lines as the individual elements of an array of strings using regular expressions to Match. The total length of the multiline string may vary as well as the text in the string. The string can contain letters, numbers and special characters. I've attached an example VI. In the example of VI, I want to only return lines starting by "device #" in the table. The number of lines starting by "device #" can vary, but I want to capture them all.
Or is there a better functioning to be used instead of the corresponding regular Expression that will give me the desired result?
Corresponding regular expression works well for this.
Ben64
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Another question about regular expressions with String.matches
don't match String.matches () method expressions when a substring of the string matches, or must match the whole string? So if I have the string '123ABC', and I ask match "1 or more letters" will be it fail because there are other that the letters in the string, but then spend if I add "1 or more letters AND numbers 1 or more? Thus, in the second case each character in the string is recorded in the research, as opposed to the first. Is that correct, or are there ways to JUST matching a substring in the string instead of all this? I'll do some examples too... but that makes sense?It must match the entire string. Use Matcher.find () to match on just a sub-string)
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Need help with regular Expression (RegEx)
Try to wrap your head around a regular Expression for the following format example: 0022-C-4452 OR 0022-C-4452-C
* The 4 digits are always numbers
* The last 5 digits are alpha numeric
* Last (if used) digit is always 'C' (in reference to the second structure)
Hold the dashes for "auto fill" if possible? This would be in the Custom Format? Sequence of keys? Or Validation? I appreciate any help!
I still think that you did not correctly describe your problem.
1 figures of 4 characters.
Optional separator.
Then 6 alphanumeric characters and ' - '.
OR
1 figures of 4 characters.
Optional separator.
Then 6 alphanumeric characters and ' - '.
Optional separator.
Character 'C '.
Note that the "-" is not a number and not an alphabetic character. It is a white space character.
Try:
function {MyRe (cString)
var cFormatted = "";
var RE_MyCode0 = /^(\d{4})[-.]) {0,1} ([A-Za-a0 - 9-] {6}) $/;
var RE_MyCode1 = /^(\d{4})[-.]) {0,1} ([A-Za-a0 - 9-] {6}) [-.] {0,1} ([C]) $/;
If (RE_MyCode0.test (CString) == true) {}
cFormatted = RegExp. $1 + '-' + RegExp. $2;
}
If (RE_MyCode1.test (CString) == true) {}
cFormatted = RegExp. $1 + '-' + RegExp. $2 + '-' + RegExp. $3;
}
Return cFormatted;
} / / end of MyRe function;some tests;
var MyString = "0022-C-4452; good channel;
Console.println ("Input:" + MyString);
Console.println ("result:" + MyRe (MyString));var MyString = "0022-C-4452-C; string of Goo;
Console.println ("Input:" + MyString);
Console.println ("result:" + MyRe (MyString));var MyString = "A022-C-4452" / / bad string;
Console.println ("Input:" + MyString);
Console.println ("result:" + MyRe (MyString));var MyString = '0022-4452-CZ' / / bad string;
Console.println ("Input:" + MyString);
Console.println ("result:" + MyRe (MyString)); -
String format using regular expressions
Input string output format... SELECT q'<select ab_c "ABC", efg "EFG" from dual>' str FROM DUAL Output: STR ------------------------------------- select ab_c "ABC", efg "EFG" from dual Required output format using regular expression... STR ------------------------------------- select 'ab_c' "ABC", 'efg' "EFG" from dual
Regular expressions have many limitations as tools of analysis, and you specify the rules you want. This expression puts quotation marks around a non-empty string before a quoted string:
SELECT regexp_replace(q'
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Search for a string using "Game Plan" or "Regular Expression to Match."
Hello
I would use the 'game plan' or the vi "Expression regular game" simply because the products that provide these vi. The result that interests me is the substring 'after '. I want to be able to specify a "substring" and get everything after the substring of the input string. However, I'm getting all confused and/or watered upward when it comes to "regular expressions". Is there a way to create a "regular expression" which acts as a 'substring' to find within the input string?
The substring is a path of partial directory that contains a colon, backslashes, etc. which are part of a directory path. If some how the "regular expression" entry must ingnore all special characters and simply to understand if the substring in the string entry and give me 'all things' after the substring in the output of 'after the string.
Use Regular Expression Match instead of match pattern; It's better.
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regular expression to exclude spaces in the string
Now that I'm that sleazy, I need help with a regular expression.
They must pass (or permutations it):
[00123]
[R012]
[r083]
These should fail (or permutations it):
[test]
[I was here]
[098 876]
Ooh [29873 eifvuh]]
I have most of it (the ' [' must be first; the '] "must be the last;) There must be some middle numbers, but I can't write an expression that will fail if there are 1 or more spaces in the string.
Thank you!
Tay
Several possibilities: try this regular expression.
^\[[a-za-Z]*[0-9]+[a-zA-Z0-9]*\]$
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I haven't used regular expressions before, and I can't find a regular expression to extract a subset of the string between two markers.
The chain;
Stuff of header I want
Stuff of header I want
Stuff of header I wantStuff of header I want
Stuff of header I want
Stuff of header I want6 ERRORS
Info I want to line 1
Info I want line 2
Info I want line 3
Info I want to line 4
Info I want to line 5
Info I want line 6
END_ERRORSFrom the string above (it is read from a text file), I try to extract the subset of string between ERRORS 6 and END_ERRORS. The number of errors (6 in this case) can be any number from 1 to 32, and the number of lines I want to extract will correspond with this number. I can provide this number of a caller VI if necessary.
My current solution, which works, but is not very elegant;
(1) using Match Regular Expression for the return of the string after you have synchronized the 6 ERRORS
(2) uses the Regular Expression matches to return all characters before game END_ERRORS of the string returned by (1)
Is there a way this can be accomplished using 1 Regular Expression Match? If so someone could suggest how, as well as an explanation of the work of the given regular expression.
Thank you very much
Alan
I used a character class to catch any word or whitespace characters. This put inside parentheses a substring matching the criteria that you can get by developing the node for regular expression matching. The \d matches the numbers and the two * s repetition of the previous term. So, \d* will find the '6', as well as "123456".
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Regular expression for middle of string
Little background. I get the data (only the numbers) and text (can contain numbers) series. Since I don't know which I get and when it starts / ends, I do the sender send 'textSTART ###textSTOP' where ' # ' is what I want to extract. # can contain text, numbers, new line, carriage return or whatever.
The same for the data : "dataSTART ###dataSTOP", where # contains only numbers.
I think I should use the match pattern, but I don't know how to make my regular expression.
Any help is appreciated.
There are many solutions if you want to only extract the string between your specific delimiters. Here's a solution:
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Oracle regular expressions - splits the string into words for
Hello
Nice day!
My requirement is to split the string into words.
So I need to identify the new line character and the semicolon (;), comma and space like terminator for string entry.
Please note that I am currently embedded blank and the comma as separator, as shown below.
Select regexp_substr('test)
TO
string in words, "([^, [: blanc:]] +) (', 1, 1) double;"How to integrate the semicolons and line break characters in regular expression Oracle?
Please notify.
Thanks and greetings
Sree
This has nothing to do with REGEXP. Is SQL * more parser does not not a semicolon at the end of the line:
SQL > select ' testto, mm\;
ERROR:
ORA-01756: city not properly finished chainSQL >
Just break the chain:
SQL > select regexp_substr ('testto, mm\;' |) '
2 string into words
3 \w+',1,level ',') of double
4. connect by level<= regexp_count('testto,mm\;'="" ||="">=>
5 string in words
6 ','\w+')
7.REGEXP_SUBSTR ('TESTTO, MM\;' |') STRINGIN
--------------------------------------
Testto
mm
string
in
WordsSQL >
Or modify SQL * more the character of endpoints:
SQL > set sqlterm.
SQL > select regexp_substr ('testto, mm\;)
2 string into words
3 \w+',1,level ',') of double
4. connect by level<=>=>
5 string in words
6 ','\w+')
7.REGEXP_SUBSTR ('TESTTO, MM\;) STRINGINTOWO
--------------------------------------
Testto
mm
string
in
WordsSQL >
SY.
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The following regular expression separates simple values separated by commas (SELECT regexp_substr (: pCsv,'[^,] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,] +', 1, level) IS NOT NULL); Exampple: 300100033146068, 300100033146071, 300100033146079 returns 300100033146068 300100033146071 300100033146079
This works very well if we use the regex with SQL IN operator select * from mytable where t.mycolumn IN (SELECT regexp_substr (: pCsv,'[^,] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,] +', 1, level) IS NOT NULL);
But this query does not work if the comma-separated value is a single literal string 'one', 'two', 'three '.
Thanks for your reply. my request was mainly on regexp_substr. Need to request was simple: any table with a column of varchar type can be used. Next time I'll give you an example.
All ways working answer for my question is is SELECT regexp_substr (: pCsv,'[^, "] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,"] +', 1, level) IS NOT NULL
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Regular expression for the format: 000-000 - 000000 000000
Hi guys,.
I need to validate the columns in a regular expression for the format of 000 000 - 000000 000000.
For example - if the column contains a value such as "500 110 - 500044 000100" then it should return 'true' otherwise 'false. '
Your timely help is well appreciated.
Thanks in advance.
Hello
inDiscover wrote:
Hi guys,.
I need to validate the columns in a regular expression for the format of 000 000 - 000000 000000.
For example - if the column contains a value such as "500 110 - 500044 000100" then it should return 'true' otherwise 'false. '
Your timely help is well appreciated.
Thanks in advance.
If you want a regular expression
REGEXP_LIKE (str
{{' ^ {\d{3} \d{3}-\d{6} \d{6}$'
)
According to your needs.
You don't need regular expressions for this. It will be more efficient to use
TRANSLATE (str
'012345678'
'999999999'
) = 999 999 - 999999 999999'
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regular expression format?
My question hopefully get me flamed, but I have to ask. When adding regular expressions newspaper Pattern Match in a configuration property is as the regular expected perl or java as expression format?
for example, in a box, I entered \bStack_Trace\b/. simply the meaning match any pattern bordered by Stack_Trace?
do I need... / / game characters or make just hack away and hyperic will add the rest?
or
do I need to get stuff like this myString.matches("\bStack_Trace\b")
Thanks in advance,
DeebohSure you can't get flamed for something not well documented nor frequent ;)
It's just a small wrapper around java.util.pattern:
http://svn.Hyperic.org/branches/HQ_3_0/src/org/Hyperic/HQ/product/StringMatcher.Java?root=Hyperic+HQ&view=markupWe should probably add a main() method here for easy test, but you can test log_track from the command line like so:
./JRE/bin/Java-Dplugins.include=system - Dlog debug = - Dlog_track.debug = true.
-jar pdk/lib/hq-product.jar t 'File server file' m - one track log.
-End = log/test.log '-Dservice.log_track.pattern=\bStack_Trace\b'Then another term:
echo "this is a Stack_Trace for deeboh" > log/test.log (will be)
echo "this is a HayStack_Trace for dougm" > log/test.log (will not match) -
regular expression - get the numbers from a string
Hello world
I'm trying to use regular expressions to get all the numbers in a string. The only problem is that the chain can vary.
For example:
It's my rope 3 and 8 I want 2 get out of those 7 numbers
Random text 9 with 5 for everyone weekend 8
How can I do?
Thanks in advance :)
Quote:
Posted by: Scott Stroz
#numbersOny #. Who will be only to crush all the numbers in a large number.
The attached code will return a nice list of numbers.
Though the numbers may contain commas or decimal points, the code can easily be adjusted -
How to capture multiple line String using regular expressions?
Hello
I have a simple program like this:
What I want to accomplish is to capture everything between > start and > to end with a single regular expression matching node. It seems that the definition of multiples? true or False does not help.
I'm using LabVIEW 2012.
If it is impossible to capture using a single node, that's fine. But I want to assure you that I can make full use of this node without combining several others.
Thank you!
> start([\w\s]*) > end
A point matches any character except line break characters. You have two of them.
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